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This is the entire proof for the 0/0 case and I am very lost. I understand choosing $r>A$, but I don't understand why we're choosing $q$. I can see why 18 is true, but I do not understand how the strict inequality becomes a weak inequality going from 18 to 19. Finally, and perhaps most importantly, I don't understand how 19 shows that $f(x)/g(x)\rightarrow A$ as $x\rightarrow a$. Can someone help me follow this through? |
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That's not the entire proof. It's half of it; the other half is on the next page (although that's definitely not clear from the way it's written). What Rudin is doing is choosing two points $p,q$ with $p<A<q$, and proving that $p<f(x)/g(x)<q$ for $x$ close enough to $a$. The use of $r$ (and other similar choices in the proof) is usually due to the fact that the theorems he is quoting require open intervals. The inequality thing from (18) to (19) is the well-known fact that if you have $x_n<k$ and $x_n\to x$, then $x\leq k$ (there might not be strict inequality: for instance take $x_n=1-1/n$, $k=1$). |
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